This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the calculations for the capacitor: Step 1: Identify the given values and convert them to SI units. • Number of parallel plates, N = 7 • Thickness of mica sheets, d = 0.5 \, mm = 0.5 × 10^-3 \, m • Relative permittivity of mica, _r = 6 • Area of each plate, A = 500 \, cm^2 = 500 × (10^-2)^2 \, m^2 = 0.05 \, m^2 • Voltage supply, V = 400 \, V • Permittivity of free space, _0 = 8.85 × 10^-12 \, F/m Step 2: Determine the number of effective dielectric layers. For a capacitor with N parallel plates, there are N-1 effective dielectric layers. Number of effective layers, n = N-1 = 7-1 = 6. Step 3: Calculate the capacitance of the capacitor. The formula for the capacitance of a multi-plate capacitor is: C = (N-1) (_r _0 A)/(d) Substitute the values: C = 6 × 6 × 8.85 × 10^-12 \, F/m × 0.05 \, m^20.5 × 10^-3 \, m C = 6 × 2.655 × 10^-12 \, F0.5 × 10^-3 C = 6 × 5.31 × 10^-9 \, F C = 31.86 × 10^-9 \, F C = 31.86 \, nF The capacitance of the capacitor is 31.86 nF. Step 4: Calculate the charge in the capacitor. The formula for charge is Q = CV. Substitute the calculated capacitance and the given voltage: Q = 31.86 × 10^-9 \, F × 400 \, V Q = 12744 × 10^-9 \, C Q = 1.2744 × 10^-5 \, C The charge in the capacitor is 1.2744 × 10^-5 \, C. Send me the next one 📸